# How to show non consequence / logical consistency?

How would I show non-consequence with this set? I can't seem to figure out the significance of the Universal and Existential indicators here.
(∀x)(∃y)Fxy
(∃y)(∀x)Fxy

and the other question i'm stumped on-
showing consistency with this set:
(∀x)(Fx → Gx)
-(∃x)(Fx & Gx)

• I'm voting to close this question because it looks like homework and no research effort is shown. – Keelan Feb 5 '15 at 12:43
• (∀x)(Fx → Gx) is not sufficient to show -(∃x)(Fx & Gx). Though they are compossible if (∀x)~(Fx) – virmaior Feb 5 '15 at 13:21
• @Keelan Something being a homework problem is not, in itself, reason to close (and no research effort can be, even if something isn't a homework problem). It can be a valid personal reason to not answer a question, but it isn't an official standard of SE or Phil SE. – Chris Sunami Feb 5 '15 at 17:30
• @ChrisSunami the problem is that questions should be answered in a way that helps the OP, while that's impossible if no effort is shown, and it therefore falls under the category questions that are unclear, too broad, or otherwise problematic to identifying the problem in a way that can be properly addressed by answerers. If you wish to discuss this in more depth, please ask it on Meta - this is a general issue and not specifically related to this question. – Keelan Feb 5 '15 at 17:42
• @Keelan That was a statement, not a question, it was in full accordance with the references you just cited, and it was in specific response to your comment to this OP. – Chris Sunami Feb 5 '15 at 17:58

I think that you are asking for :

how to prove that (∃y)(∀x)Fxy is not a logical consequence [see this post for the definition] of (∀x)(∃y)Fxy.

If so, in order to prove it, we have to find a counter-example, i.e. an interpretation such that (∀x)(∃y)Fxy is true while (∃y)(∀x)Fxy is false.

The "standard" counter-example is found assuming as domain for the interpretation the set N of natural numbers and with the relation < ("less-then") as interpretation for the binary predicate symbol F.

We have that in N it is true that :

(∀x)(∃y)(x < y)

because for every natural number n it is enough to choose n+1 and we have n < n+1.

But :

(∃y)(∀x)(x < y)

is false, because there is no number which is greater than all other numbers.

Regarding the second problem, it amounts to show that :

(∀x)(Fx → Gx) and ¬(∃x)(Fx & Gx) are simultaneously satsfiable.

We can show it applying some simple transformations.

1) A → B is equivalent to ¬A ∨ B

2) ¬(A & B) is equivalent to ¬A ∨ ¬B [De Morgan]

3) ¬(∃x)A is equivalent to (∀x)¬A.

Consider now : (∀x)(Fx → Gx); by 1) it is equivalent to : (∀x)(¬Fx ∨ Gx).

Consider : ¬(∃x)(Fx & Gx); by 3) it is equivalent to : (∀x)¬(Fx & Gx) and by 2) to : (∀x)(¬Fx ∨ ¬Gx).

Thus the problem is equivalento to show that :

(∀x)(¬Fx ∨ Gx) and (∀x)(¬Fx ∨ ¬Gx) are simultaneously satsfiable.

We can prove this assuming a domain with only one black ball and interpret the two predicate symbols Fx and Gx as "x is white" and "x is square" respectively.

With this interpretation, the first formula : (∀x)(¬Fx ∨ Gx) means :

"all objects in the domain are not-white and square"

while the second formula : (∀x)(¬Fx ∨ ¬Gx) means :

"all objects in the domain are not-white and not-square".

Due to the fact that in the domain there is only one black ball, i.e. a not-white ball, both disjunctions are satisfied simultaneously.

Mauro's answer is correct. In the case, however, that your main difficulty is conceptualizing the quantifiers, you might try translating into natural language:

For the first, let's take x and y and people and F as "loves".

(∀x)(∃y)Fxy For all people, there exists a person he or she loves or Everybody loves someone. (∃y)(∀x)Fxy There exists a person who is loved by all people.

Clearly, those two are not the same, and neither follows from the other.

For the second one, we can render F as "has an infinite amount of money" and G as "is happy".

(∀x)(Fx → Gx) It is true for everyone that if you have an infinite amount of money, you are happy -(∃x)(Fx & Gx) It is not true that there is a person who has an infinite amount of money and is happy

These can both be true just in the case that there is no person that has an infinite amount of money. (∀x)(-Fx)

• Thank you for the response. That was actually very helpful, I was struggling to conceptualize it in natural language. I'm still sort of confused about how to 'answer' the question. the format is supposed to be something like: Domain: {1,2}; F: {1}; G: {2}. but for Q#2 that is not correct, because I'm still pretty lost on how i'm supposed to answer like that. For consistency, I want to make both true. So for the first part, everything is such that Fx goes to Gx. So I can just put any number in F and G? And next something is such that it is not Fx&Gx, I add something unused to the domain? – jacques Feb 6 '15 at 9:19
• so for #2, I got domain: {1}; F: {}; G: {}. I think i also could have answered- domain: {1,2,3}; F: {1}; G: {2}. So that mostly makes sense to me. for #1, however, I am pretty stumped. – jacques Feb 6 '15 at 9:37
• OK, using this format: For the first, you want to define domains such that one of those statements is true and the other is false, and objects in those domains should be ordered pairs such as (1,2) since the function takes two parameters. For the second one, you want to define domains as you are doing, but such that both statements are true. The answer you chose works, the other one does not (hint: check against the first statement). BTW, you should upvote any helpful answer (little arrow above the number). If any answer fully answers your question, accept it (checkmark). – Chris Sunami Feb 6 '15 at 14:48
• Also, to help translate between our natural language renditions and the domain format, think of each number as being the name of a person. In other words, "1" is a person, F:{(1,2), (2,1)} means 1 loves 2 and 2 loves 1. – Chris Sunami Feb 6 '15 at 14:54